Respuesta :

superman1987

Answer:

1.1724 g.

Explanation:

  • Firstly we need to calculate the percentage of pure MgSO₄ in (MgSO₄.7H₂O).
  • If we have 1.0 mol of MgSO₄.7H₂O, then we will have the the mass of its molecular mass.

The molecular mass of 1.0 mol (MgSO₄.7H₂O) = 246.4 g/mol.

The molecular mass of pure MgSO₄ = 120.366 g/mol.

The molecular mass of 7(H₂O) = 7(18.0 g/mol) = 126.0 g/mol.

∴ The mass % of pure MgSO₄ = [(mass of pure MgSO₄)/(mass of MgSO₄.7H₂O)] x 100 = [(120.366 g/mol)/(246.4 g/mol)] x 100 = 48.85%.

∴ the quantity of pure MgSO₄ = (mass of MgSO₄.7H₂O)(% of MgSO₄/100) = (2.4 g)(48.85/100) = 1.1724 g.

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